Question:
A transmitting station releases waves of wavelength $960 \mathrm{~m}$. A capacitor of $256 \mu \mathrm{F}$ is used in the resonant
circuit. The self inductance of coil necessary for resonance is $\times 10^{-8} \mathrm{H}$
Solution:
(10)
Since resonance $\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{LC}}}$
$\therefore 2 \pi f=\frac{1}{\sqrt{L C}}$
$\therefore 4 \pi^{2} \frac{C^{2}}{\lambda^{2}}=\frac{1}{L C}$
$\therefore \frac{4 \pi^{2} \times 9 \times 10^{8} \times 9 \times 10^{8}}{960 \times 960}=\frac{1}{\mathrm{~L} \times 2.56 \times 10^{-6}}$
$=\frac{375 \times 960}{10^{-6} \times 4 \times \pi^{2} \times 9 \times 10^{16}}=\frac{10^{3}}{10^{10}}$
$=10^{-7} \mathrm{H}$
$=10 \times 10^{-8}$